Algorithm Gossip: 完美数优化版_综合

package main01;

import java.util.ArrayList;
import java.util.Scanner;
/**
*
* @author 欧阳子木
*
*/
public class CoreJava {

   /**
   * .Algorithm Gossip: 完美数说明如果有一数n，其真因数（Proper factor）的总和等于n，则称之为完美数（Perfect
   * Number），例如以下几个数都是完美数：
   * 6 = 1 + 2 + 3
   * 28 = 1 + 2 + 4 + 7 + 14
   * 496 = 1 + 2 +4 + 8 + 16 + 31 + 62 + 124 + 248
   * 由上式可知 2 * 28 = 1 + 2 + 4 + 7 + 14 + 28
   * 观察上式可得
   * 2 * 28 = (2 ^ 0 + 2 ^ 1 + 2 ^ 2) * (7 ^ 0 + 7 ^ 1)
   *
   * @param args
   */
   public static void main(String[] args) {
       // TODO Auto-generated method stub
        Scanner input = new Scanner(System.in);
        System.out.println("输入一个数字");
        PerfectNumber(input.nextInt());
        input.close();
   }

   // 第一步求得一定数量的质数集合
   public static ArrayList<Integer> Eratosthenes(int number) {
       ArrayList<Integer> list = new ArrayList<Integer>();
       int i = 0;
       int j = 0;
       int array[] = new int[number + 1];
       // 初始化数组不考虑1
       for (i = 2; i <= array.length - 1; i++) {
           array[i] = 1;
       }
       for (i = 2; i <= array.length - 1; i++) {
           // 从2开始凡是为2得倍数，3的倍数，4的倍数。。。。。。则将该值设为0
           for (j = i * i; j <= array.length - 1; j += i) {
               array[j] = 0;
           }
       }
       //
       for (i = 2; i <= array.length - 1; i++) {
           if (array[i] == 1) {
               list.add(i);
           }
       }
       return list;
   }
//得到质因数集合
   public static ArrayList<Integer> isAll_factor(int number) {
       ArrayList<Integer> list = new ArrayList<Integer>();
       ArrayList<Integer> temp_List = Eratosthenes(number);
       for (int i = 0; temp_List.get(i) * temp_List.get(i) <= number;) {
           if (number % temp_List.get(i) == 0) {
               list.add(temp_List.get(i));
               number = number / temp_List.get(i);
           } else {
               i++;
           }

       }
       list.add(number);

       return list;
   }
   //因数求和,并判断是否为完美数
   public static boolean isPerfectNumber(ArrayList<Integer> list, int number) {
       int i = 0;
       int temp2 = 1;
       int temp3 = 1;
       int sum = 1;
       //思想：2 * 28 = (2 ^ 0 + 2 ^ 1 + 2 ^ 2) * (7 ^ 0 + 7 ^ 1)
       while (i < list.size()) {
           do {
               temp2 *= list.get(i);
               temp3 += temp2;
               i++;
           } while (i < list.size() - 1 && list.get(i - 1) == list.get(i));
           sum *= temp3;
           temp2 = 1;
           temp3 = 1;
       }
       if (sum == 2 * number) {
           return true;
       }
       return false;
   }
   //重复调用该方法,判断指定范围内的完美数，并输出其个数
   public static void PerfectNumber(int number){
       int count = 0;
       for(int i = 2 ; i <= number; i++){
              if(isPerfectNumber(isAll_factor(i), i)){
                  System.out.println(i + "是个完美数");
                  count++;
              }
       }
       System.out.println("完美数个数为:" + count + "个");
   }



}